Topic 3.3 key terms
- derivative of an inverse function
- If f is differentiable and one-to-one, f(a) = b and f′(a) ≠ 0, then (f⁻¹)′(b) = 1/f′(a): find the input that gave the output, then take the reciprocal of the slope there.
- reciprocal slopes at mirrored points
- Reflecting a graph across y = x swaps each point (a, b) to (b, a) and swaps the rise and run of every tangent triangle, so the slopes there multiply to 1.
- provided the derivative exists
- Where f has a horizontal tangent (f′(a) = 0), its inverse has a vertical tangent at b = f(a) and no derivative there, as with x³ and ∛x at 0, or arcsin at ±1.
- derivative of arcsin
- d/dx arcsin x = 1/√(1 − x²) for −1 < x < 1; with the chain rule, d/dx arcsin(u) = u′/√(1 − u²).
- derivative of arccos
- d/dx arccos x = −1/√(1 − x²) for −1 < x < 1; with the chain rule, d/dx arccos(u) = −u′/√(1 − u²).
- derivative of arctan
- d/dx arctan x = 1/(1 + x²) for every real x; with the chain rule, d/dx arctan(u) = u′/(1 + u²).
- angle of a tracking camera
- A camera d away from a vertical climb at height h has angle θ = arctan(h/d), so θ′ = d·h′/(d² + h²) radians per second.
Transcript
1,377 words · 4 parts
Six kilometers away
It’s two forty-nine in the morning on the second of March, 2019, and across the lagoon a Falcon 9 is about to launch Crew Dragon on its very first trip to orbit. We’re on the shore next to a tracking camera, six kilometers from the pad, and its one job tonight is to keep that rocket right in the middle of the picture, and before long, that job will be yours. Here it goes.
From six kilometers away you see the flame long before you hear it, because the sound takes about seventeen and a half seconds to cross the water. There it is. As the rocket climbs, the camera tilts up to follow it, and at first it has to tilt faster and faster to keep up. But keep your eye on these two numbers, because about forty seconds in, the rocket is still speeding up and the camera starts turning more slowly.
By ninety seconds the rocket is climbing three times as fast as it was at forty, and the camera is turning at about a third of its top speed. By the end of this lesson, you’ll be able to work out exactly how fast this camera has to turn at any second of the flight, and why it slows down while the rocket speeds up. Then you’ll set its motor yourself and see if the rocket stays in the frame.
The camera’s angle is an inverse function called arctangent, so first we need derivatives of inverses. The rocket can teach us those with a simpler question, how long each kilometer of the climb takes.
Seconds per kilometer
Let’s put a glowing ring at every kilometer of height and check the clock each time the rocket passes one. The first kilometer takes about twenty seconds, because the rocket starts out slow, and the second takes only nine.
By eight kilometers, a whole kilometer goes by in under four seconds, and right at sixty seconds the rocket is climbing two hundred seventy-nine and a half meters every second. At exactly that rate, how many seconds would the next kilometer take? Give your answer to the nearest tenth of a second.
We just turned a rate upside down, because zero point two eight kilometers every second and three point six seconds per kilometer describe the very same speed. That pace belongs to a second function hiding in this flight. The altitude h of t turns a time into a height, and its inverse, T of y, turns a height y into the time the rocket gets there.
The derivative of that inverse, dT dy, is exactly the pace, seconds per kilometer, which is one over the climb rate. Pause any time and drag this marker up the column, and the world will tell you when the rocket gets there and its pace there, in seconds per kilometer. On this glass card the climb is a graph of height against time, and the tangent at sixty seconds rises zero point two eight kilometers for every second.
Let’s flip the graph over so that height runs along the bottom. The triangle turns with it, the rise and the run trade places, and the slope turns upside down, into about three point six seconds per kilometer. That flip is the whole idea of this lesson, so let’s take it to the board, with numbers that are easier to see.
The mirror
Let’s use f of x equals x squared, but only for x of zero or more, so every output comes from exactly one input. At x equals two, f gives us four, and the slope there is two times two, so the tangent rises four for every one it runs.
In a moment we’ll reflect this graph across the line y equals x, which sends two comma four to four comma two. Before we do, make a prediction. What will the slope of the mirror image be at four comma two?
Let’s reflect it. Every point swaps its coordinates, and the triangle folds over with it, so now it runs four and rises one. The mirror image is the square root of x, and its slope here is one fourth.
The two slopes multiply to exactly one, four times a quarter. We can check it with the power rule, too. The derivative of root x is one over two root x, and at four that’s one fourth, exactly what the mirror said. The rule, then, is to find the input that gave you this output and flip the slope at that input.
In symbols, the slope of f inverse at b is one over f prime of a, where a is the input that f sends to b. In our picture, a is two and b is four.
To see why it always works, call the inverse g. Then f of g of x gives back x every time, and we can differentiate both sides with the chain rule. The left side becomes f prime of g of x times g prime of x, the right side becomes one. Dividing gives g prime of x equals one over f prime of g of x.
That’s the same rule, because g of x is exactly the input that f sends to x. That only works when the bottom isn’t zero. Take x cubed, which has a flat tangent at zero. Its mirror image, the cube root, stands straight up there, and a vertical line has no slope, so the cube root has no derivative at zero. That’s why the rule comes with the words provided the derivative exists, because wherever f is flat, its inverse is vertical.
Find the input first
On the AP exam you’ll rarely get a formula for the inverse. You’ll get a table like this one, for an increasing, differentiable function f, with g as its inverse. Let’s find g prime of four in two separate moves, because doing it in one is where almost everybody slips.
First I find the input that gave us this output. Four sits in the f of x row, right below x equals two, so g of four is two. Then I flip the slope at that input, and since f prime of two is two point five, g prime of four is one over two point five, or zero point four.
The tempting mistake is to use f prime of four, because four is in the question, but four is an output of f, and f prime needs an input. As a sanity check, g’s average slope is a half on the stretch just before four and a third on the stretch just after, so a slope of zero point four fits right in.
Now you try one with the same table. What’s g prime of seven? You can type it as a fraction, or as a decimal to three places.
With a formula it’s the same two moves. Take f of x equals x cubed plus two x minus one, and let’s find g prime of two and the tangent line to g there. First we need the input that f sends to two. Solving a cubic sounds awful, but questions like this are usually built around an easy input, so let’s try one first, and one works, because one plus two minus one is two.
f prime of x is three x squared plus two, which is five at x equals one, so g prime of two is one fifth. The point two comma one is on g, because f of one is two, so the tangent line is y minus one equals one fifth times x minus two.
When a free-response question asks you to justify this, it wants two reasons, why the inverse exists and why its slope is defined at two, and this is how they read. f prime of x is three x squared plus two, which is always positive, so f is increasing and has an inverse. Since f of one is two and f prime of one is five, not zero, g is differentiable at two, and g prime of two is one fifth.
Try this one on your own. If f of x is x to the fifth plus x, what’s the slope of f inverse at thirty-four?